[proofplan] We apply the bracket identity from part (ii) of the [Itô Isometry](/theorems/2092) twice in succession. First, $\langle H \cdot M, K \cdot N \rangle_t = \int_0^t H_s \, d\langle M, K \cdot N \rangle_s$ (treating $K \cdot N$ as the second martingale). Then the inner bracket $\langle M, K \cdot N \rangle$ is itself expressed as $\int_0^{(\cdot)} K_s \, d\langle M, N \rangle_s$ by applying part (ii) again (with the roles of integrand and martingale swapped). Substituting and using the associativity of Lebesgue–Stieltjes integration against the measure $d\langle M, N \rangle_s$ produces the result. [/proofplan]

[step:Express $\langle H \cdot M, K \cdot N \rangle_t$ using the bracket identity for $H \cdot M$] Both $H \cdot M$ and $K \cdot N$ belong to $\mathcal{M}^2_c$ by the [Itô Isometry](/theorems/2092) (part (i)), since $H \in L^2(M)$ and $K \in L^2(N)$. By part (ii) of the [Itô Isometry](/theorems/2092), applied with the continuous square-integrable martingale $H \cdot M$ and the test martingale $K \cdot N \in \mathcal{M}^2_c$, \begin{align*} \langle H \cdot M, K \cdot N \rangle_t = \int_0^t H_s \, d\langle M, K \cdot N \rangle_s. \end{align*} [/step]

[step:Compute the inner bracket $\langle M, K \cdot N \rangle_s$ using the bracket identity for $K \cdot N$] By symmetry of covariation ([Properties of Covariation](/theorems/2086), part (ii)), $\langle M, K \cdot N \rangle = \langle K \cdot N, M \rangle$. Applying part (ii) of the [Itô Isometry](/theorems/2092) with integrand $K$, integrator $N$, and test martingale $M$, \begin{align*} \langle K \cdot N, M \rangle_s = \int_0^s K_u \, d\langle N, M \rangle_u = \int_0^s K_u \, d\langle M, N \rangle_u, \end{align*} where the second equality uses symmetry $\langle N, M \rangle = \langle M, N \rangle$. [/step]

[step:Substitute and identify the iterated integral as $\int_0^t H_s K_s \, d\langle M, N \rangle_s$]The measure $d\langle M, K \cdot N \rangle_s$ is the Lebesgue–Stieltjes measure of the function $s \mapsto \langle M, K \cdot N \rangle_s = \int_0^s K_u \, d\langle M, N \rangle_u$. By the definition of the Lebesgue–Stieltjes integral, this measure has density $K_s$ with respect to $d\langle M, N \rangle_s$: for any Borel set $B \subseteq [0, \infty)$, \begin{align*} \int_B d\langle M, K \cdot N \rangle_s = \int_B K_s \, d\langle M, N \rangle_s. \end{align*} Substituting into the expression from the first step, \begin{align*} \langle H \cdot M, K \cdot N \rangle_t = \int_0^t H_s \, d\langle M, K \cdot N \rangle_s = \int_0^t H_s K_s \, d\langle M, N \rangle_s. \end{align*}[/step]

[guided]The key idea is that the bracket identity in part (ii) of the [Itô Isometry](/theorems/2092) can be applied in two different ways: once to "peel off" the integrand $H$ from $H \cdot M$, and once to "peel off" the integrand $K$ from $K \cdot N$. Starting from the bracket $\langle H \cdot M, K \cdot N \rangle_t$, part (ii) of the [Itô Isometry](/theorems/2092) gives \begin{align*} \langle H \cdot M, K \cdot N \rangle_t = \int_0^t H_s \, d\langle M, K \cdot N \rangle_s. \end{align*} Now the inner bracket $\langle M, K \cdot N \rangle$ involves a stochastic integral $K \cdot N$. By symmetry of covariation ([Properties of Covariation](/theorems/2086)), $\langle M, K \cdot N \rangle = \langle K \cdot N, M \rangle$. Applying part (ii) of the [Itô Isometry](/theorems/2092) again, now with integrand $K$, integrator $N$, and test martingale $M$: \begin{align*} \langle K \cdot N, M \rangle_s = \int_0^s K_u \, d\langle N, M \rangle_u = \int_0^s K_u \, d\langle M, N \rangle_u. \end{align*} What does this mean for the measure $d\langle M, K \cdot N \rangle_s$? The function $s \mapsto \langle M, K \cdot N \rangle_s$ is a Lebesgue–Stieltjes integral of $K$ against $d\langle M, N \rangle$. Therefore the Radon–Nikodym derivative of the signed measure $d\langle M, K \cdot N \rangle$ with respect to $d\langle M, N \rangle$ is $K_s$. This is the chain-rule property of Lebesgue–Stieltjes integrals: if $A(t) = \int_0^t f(s) \, d\mu(s)$, then $\int g \, dA = \int g \, f \, d\mu$. Applying this with $g = H$, $f = K$, $\mu = d\langle M, N \rangle$: \begin{align*} \langle H \cdot M, K \cdot N \rangle_t = \int_0^t H_s \, d\langle M, K \cdot N \rangle_s = \int_0^t H_s K_s \, d\langle M, N \rangle_s. \end{align*}[/guided]